Problem: Graph this system of equations and solve. $-14x+2y = 4$ $6x-2y = 4$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $-14x+2y = 4$ , to slope-intercept form. $y = 7 x + 2$ The y-intercept for the first equation is $2$ , so the first line must pass through the point $(0, 2)$ The slope for the first equation is $7$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move up $1$ position to the right. $7$ positions up from $(0, 2)$ is $(1, 9)$ Graph the blue line so it passes through $(0, 2)$ and $(1, 9)$ Convert the second equation, $6x-2y = 4$ , to slope-intercept form. $y = 3 x - 2$ The y-intercept for the second equation is $-2$ , so the second line must pass through the point $(0, -2)$ The slope for the second equation is $3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up $1$ position to the right. $3$ positions up from $(0, -2)$ is $(1, 1)$ Graph the green line so it passes through $(0, -2)$ and $(1, 1)$ The solution is the point where the two lines intersect. The lines intersect at $(-1, -5)$.